POJ 2068 Nim#双人dp博弈

Posted 2020-06-18 ATM

http://poj.org/problem?id=2068

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;

int dp[25][(1<<13)+5];//dp[i][j]表示轮到第i个人取时，剩j个石头
int n,s,m[25];

int DFS(int pos,int remain)
{
    if(dp[pos][remain]!=-1)
        return dp[pos][remain];//记忆化
    if(remain==0)//必胜态
        return dp[pos][remain]=1;

    dp[pos][remain]=0;
    for(int i=1;i<=min(m[pos],remain);i++)
        if(!DFS((pos+1)%n,remain-i))//对方为必败态的话，己方则必胜
            dp[pos][remain]=1;
    return dp[pos][remain];
}

int main()
{
    while(scanf("%d",&n)&&n)
    {
        scanf("%d",&s);
        n*=2;
        for(int i=0;i<n;i++)
            scanf("%d",&m[i]);
        memset(dp,-1,sizeof(dp));
        printf("%d\n",DFS(0,s));
    }
    return 0;
}